Hello,

Firebird 2.0.3.12981.0-r6 on Gentoo Linux. I have a table called
AUSZUEGE and would like to create a unique index on some of the
columns of that table. But it wouldn't let me - there are still
some duplicate rows in that table (which I try to avoid in the
future by creating that unique index). How can I find that dupli-
cate rows?

I try

CREATE UNIQUE INDEX AUSINDEX
on AUSZUEGE (BLZ, KONTO, AUSZUGNR, NEUGUTDATUM);

(with BLZ=CHAR(8), KONTO=VARCHAR(10), AUSZUGNR=INTEGER, and
NEUGUTDATUM=DATE)

and get
"attempt to store duplicate value (visible to active transactions)
in unique index "AUSINDEX"."

So I need to identify those duplicate values by something like
SELECT * from AUSZUEGE where COUNT(two_consecutive_identical_rows)>1;
but how can I say "two_consecutive_identical_rows" in SQL?

I already thought of adding a temporary VARCHAR column composed
of the string representation of the desired index fields and get
an output ORDER BY that colum - then the two identical rows would
already be consecutive. But how to tell FB just to show rows where
there _are_ at least two identical consecutive rows?

Thank you,

Matt


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